Question

When solutions of barium chloride (aq) and sodium sulfate (aq) are mixed, what is/are the spectator ion(s)?

Answer 1

Double displacement reactions may be defined as the chemical reactions in which one component each of both the reacting molecules is exchanged to form the products. During this reaction, the cations and anions of two different compounds switch places, forming two entirely different compounds.
The general equation which represents a double displacement reaction can be written as:
$AB + CD \to AD + CB$
Double displacement reactions generally take place in aqueous solutions in which the ions precipitate and there is an exchange of ions.

Complete answer:
On mixing a solution of barium chloride $(BaC{l_2})$ with sodium sulfate $(N{a_2}S{O_4})$, a white precipitate of barium sulphate is immediately formed. These reactions are ionic in nature. The reactants change into ions when dissolved in water and there is an exchange of ions in solution. This results in the formation of product molecules.
the equation for the formation of $BaS{O_4}(s)$ is as follows:
$BaC{l_2}(aq) + N{a_2}S{O_4}(aq) \to BaS{O_4}(s) \downarrow + 2NaCl(aq)$
It can be rewritten as
$B{a^{2 + }}(aq) + 2C{l^ - }(aq) + 2N{a^ + }(aq) + SO_4^{2 - }(aq) \to BaS{O_4}(s) \downarrow + 2N{a^ + }(aq) + 2C{l^ - }(aq)$
So, $N{a^ + }$and $C{l^ - }$ are found in solution both before and after the reaction and hence play no part in macroscopic reaction and therefore are spectator ions.

ADDITIONAL INFORMATION: Alternatively, we could write the net ionic equation:
$B{a^{2 + }}(aq) + SO_4^{2 - }(aq) \to BaS{O_4}(s) \downarrow$

Note:
A spectator ion is an ion that does not take part in the chemical reaction and is found in solution both before and after the reaction. And to determine spectator ions, compare the reactant and product sides of the rewritten reaction and cross out the spectator ions. Any dissolved ions that appear in the same form on both sides are spectator ions.